[vtkusers] Problems using vtkConnectivityFilter on Image Data

[David Gobbi]

Hi Bill,

Yes, I intend to do that one of these days.  The vtkMRIBrainExtractor filter
in particular is a very useful tool for visualization, but it really doesn't
belong in VTK as a core class.  Remote modules are the perfect solution
for such classes.

 - David


On Fri, Nov 6, 2015 at 1:38 PM, Bill Lorensen <bill.lorensen at gmail.com>
wrote:

> David,
>
> Should you add your ImageSegmentation as an External Module?
>
> Bill
>
> On Fri, Nov 6, 2015 at 12:49 PM, David Gobbi <david.gobbi at gmail.com>
> wrote:
> > Hi Darya,
> >
> > The vtkConnectivityFilter is meant for use on mesh data.  It suspect
> > that when you used it on image data, it caused your computer to run
> > out of memory.
> >
> > I have a connectivity filter that I wrote specifically for images, but it
> > is on github.  I haven't contributed it to VTK yet.  But you can try it
> > if you are interested:
> >
> > https://github.com/dgobbi/AIRS/tree/master/ImageSegmentation
> >
> > You can also try the vtkImageThresholdConnectivity filter that comes
> > with VTK:
> >
> http://www.vtk.org/doc/nightly/html/classvtkImageThresholdConnectivity.html
> >
> >  - David
> >
> >
> >
> > On Fri, Nov 6, 2015 at 10:36 AM, Darya Yelshyna via vtkusers
> > <vtkusers at vtk.org> wrote:
> >>
> >> Hi everyone!
> >>
> >> I was trying to apply vtkConnectivityFilter to vtkImageData, but the
> >> program crashes at this step.
> >> "This application has requested the Runtime to terminate it in an
> unusual
> >> way." This is this a part of my code:
> >>
> >> dilateFilter = vtk.vtkImageContinuousDilate3D()
> >> dilateFilter.SetKernelSize(5, 5, 5)
> >> dilateFilter.SetInputConnection(logic2.GetOutputPort())
> >> dilateFilter.Update()
> >>
> >> connectFilter = vtk.vtkConnectivityFilter()
> >> connectFilter.SetExtractionModeToLargestRegion()
> >> connectFilter.SetInputConnection(dilateFilter.GetOutputPort())
> >> connectFilter.Update()
> >>
> >> I don't understand what is happening.. Can anybody help me?
> >>
> >> Thank you in advance.
> >>
> >> Best regards,
> >> Darya
>
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